FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 – August 4, 1997, Plovdiv, BULGARIA First day — August 1, 1997 Problems and Solutions Problem 1. Let {ε n } ∞ n=1 be a sequence of positive real numbers, such that lim n→∞ ε n = 0. Find lim n→∞ 1 n n  k=1 ln  k n + ε n  , where ln denotes the natural logarithm. Solution. It is well known that −1 =  1 0 ln xdx = lim n→∞ 1 n n  k=1 ln  k n  (Riemman’s sums). Then 1 n n  k=1 ln  k n + ε n  ≥ 1 n n  k=1 ln  k n  −→ n→∞ −1. Given ε > 0 there exist n 0 such that 0 < ε n ≤ ε for all n ≥ n 0 . Then 1 n n  k=1 ln  k n + ε n  ≤ 1 n n  k=1 ln  k n + ε  . Since lim n→∞ 1 n n  k=1 ln  k n + ε  =  1 0 ln(x + ε)dx =  1+ε ε ln xdx 1 we obtain the result when ε goes to 0 and so lim n→∞ 1 n n  k=1 ln  k n + ε n  = −1. Problem 2. Suppose ∞  n=1 a n converges. Do the following sums have to converge as well? a) a 1 + a 2 + a 4 + a 3 + a 8 + a 7 + a 6 + a 5 + a 16 + a 15 + ···+ a 9 + a 32 + ··· b) a 1 + a 2 + a 3 + a 4 + a 5 + a 7 + a 6 + a 8 + a 9 + a 11 + a 13 + a 15 + a 10 + a 12 + a 14 + a 16 + a 17 + a 19 + ··· Justify your answers. Solution. a) Yes. Let S = ∞  n=1 a n , S n = n  k=1 a k . Fix ε > 0 and a number n 0 such that |S n − S| < ε for n > n 0 . The partial sums of the permuted series have the form L 2 n−1 +k = S 2 n−1 + S 2 n − S 2 n −k , 0 ≤ k < 2 n−1 and for 2 n−1 > n 0 we have |L 2 n−1 +k − S| < 3ε, i.e. the permuted series converges. b) No. Take a n = (−1) n+1 √ n .Then L 3.2 n−2 = S 2 n−1 + 2 n−1 −1  k=2 n−2 1 √ 2k + 1 and L 3.2 n−2 − S 2 n−1 ≥ 2 n−2 1 √ 2 n −→ n→∞ ∞, so L 3.2 n−2 −→ n→∞ ∞. Problem 3. Let A and B be real n×n matrices such that A 2 +B 2 =AB. Prove that if BA − AB is an invertible matrix then n is divisible by 3. Solution. Set S = A + ωB, where ω = − 1 2 + i √ 3 2 . We have S S = (A + ωB)(A + ωB) = A 2 + ωBA + ωAB + B 2 = AB + ωBA + ωAB = ω(BA −AB), because ω + 1 = −ω. Since det(SS) = det S. det S is a real number and det ω(BA − AB) = ω n det(BA − AB) and det(BA − AB) = 0, then ω n is a real number. This is possible only when n is divisible by 3. 2 Problem 4. Let α be a real number, 1 < α < 2. a) Show that α has a unique representation as an infinite product α =  1 + 1 n 1  1 + 1 n 2  . . . where each n i is a positive integer satisfying n 2 i ≤ n i+1 . b) Show that α is rational if and only if its infinite product has the following property: For some m and all k ≥ m, n k+1 = n 2 k . Solution. a) We construct inductively the sequence {n i } and the ratios θ k = α  k 1 (1 + 1 n i ) so that θ k > 1 for all k. Choose n k to be the least n for which 1 + 1 n < θ k−1 (θ 0 = α) so that for each k, (1) 1 + 1 n k < θ k−1 ≤ 1 + 1 n k − 1 . Since θ k−1 ≤ 1 + 1 n k − 1 we have 1 + 1 n k+1 < θ k = θ k−1 1 + 1 n k ≤ 1 + 1 n k −1 1 + 1 n k = 1 + 1 n 2 k − 1 . 3 Hence, for each k, n k+1 ≥ n 2 k . Since n 1 ≥ 2, n k → ∞ so that θ k → 1. Hence α = ∞  1  1 + 1 n k  . The uniquness of the infinite product will follow from the fact that on every step n k has to be determine by (1). Indeed, if for some k we have 1 + 1 n k ≥ θ k−1 then θ k ≤ 1, θ k+1 < 1 and hence {θ k } does not converge to 1. Now observe that for M > 1, (2)  1 + 1 M  1 + 1 M 2  1 + 1 M 4  ··· = 1+ 1 M + 1 M 2 + 1 M 3 +··· = 1+ 1 M − 1 . Assume that for some k we have 1 + 1 n k − 1 < θ k−1 . Then we get α (1 + 1 n 1 )(1 + 1 n 2 ) . . . = θ k−1 (1 + 1 n k )(1 + 1 n k+1 ) . . . ≥ θ k−1 (1 + 1 n k )(1 + 1 n 2 k ) . . . = θ k−1 1 + 1 n k −1 > 1 – a contradiction. b) From (2) α is rational if its product ends in the stated way. Conversely, suppose α is the rational number p q . Our aim is to show that for some m, θ m−1 = n m n m − 1 . Suppose this is not the case, so that for every m, (3) θ m−1 < n m n m − 1 . 4 For each k we write θ k = p k q k as a fraction (not necessarily in lowest terms) where p 0 = p, q 0 = q and in general p k = p k−1 n k , q k = q k−1 (n k + 1). The numbers p k −q k are positive integers: to obtain a contradiction it suffices to show that this sequence is strictly decreasing. Now, p k − q k − (p k−1 − q k−1 ) = n k p k−1 − (n k + 1)q k−1 − p k−1 + q k−1 = (n k − 1)p k−1 − n k q k−1 and this is negative because p k−1 q k−1 = θ k−1 < n k n k − 1 by inequality (3). Problem 5. For a natural n consider the hyperplane R n 0 =  x = (x 1 , x 2 , . . . , x n ) ∈ R n : n  i=1 x i = 0  and the lattice Z n 0 = {y ∈ R n 0 : all y i are integers}. Define the (quasi–)norm in R n by x p =  n  i=1 |x i | p  1/p if 0 < p < ∞, and x ∞ = max i |x i |. a) Let x ∈ R n 0 be such that max i x i − min i x i ≤ 1. For every p ∈ [1, ∞] and for every y ∈ Z n 0 prove that x p ≤ x + y p . b) For every p ∈ (0, 1), show that there is an n and an x ∈ R n 0 with max i x i − min i x i ≤ 1 and an y ∈ Z n 0 such that x p > x + y p . 5 Solution. a) For x = 0 the statement is trivial. Let x = 0. Then max i x i > 0 and min i x i < 0. Hence x ∞ < 1. From the hypothesis on x it follows that: i) If x j ≤ 0 then max i x i ≤ x j + 1. ii) If x j ≥ 0 then min i x i ≥ x j − 1. Consider y ∈ Z n 0 , y = 0. We split the indices {1, 2, . . . , n} into five sets: I(0) = {i : y i = 0}, I(+, +) = {i : y i > 0, x i ≥ 0}, I(+, −) = {i : y i > 0, x i < 0}, I(−, +) = {i : y i < 0, x i > 0}, I(−, −) = {i : y i < 0, x i ≤ 0}. As least one of the last four index sets is not empty. If I(+, +) = Ø or I(−, −) = Ø then x + y ∞ ≥ 1 > x ∞ . If I(+, +) = I(−, −) = Ø then  y i = 0 implies I(+, −) = Ø and I(−, +) = Ø. Therefore i) and ii) give x + y ∞ ≥ x ∞ which completes the case p = ∞. Now let 1 ≤ p < ∞. Then using i) for every j ∈ I(+, −) we get |x j + y j | = y j − 1 + x j + 1 ≥ |y j | −1 + max i x i . Hence |x j + y j | p ≥ |y j | − 1 + |x k | p for every k ∈ I(−, +) and j ∈ I(+, −). Similarly |x j + y j | p ≥ |y j | − 1 + |x k | p for every k ∈ I(+, −) and j ∈ I(−, +); |x j + y j | p ≥ |y j | + |x j | p for every j ∈ I(+, +) ∪ I(−, −). Assume that  j∈I(+,−) 1 ≥  j∈I(−,+) 1. Then x + y p p − x p p =  j∈I(+,+)∪I(−,−) (|x j + y j | p − |x j | p ) +    j∈I(+,−) |x j + y j | p −  k∈I(−,+) |x k | p   +    j∈I(−,+) |x j + y j | p −  k∈I(+,−) |x k | p   ≥  j∈I(+,+)∪I(−,−) |y j | +  j∈I(+,−) (|y j | −1) 6 +    j∈I(−,+) (|y j | −1) −  j∈I(+,−) 1 +  j∈I(−,+) 1   = n  i=1 |y i | − 2  j∈I(+,−) 1 = 2  j∈I(+,−) (y j − 1) + 2  j∈I(+,+) y j ≥ 0. The case  j∈I(+,−) 1 ≤  j∈I(−,+) 1 is similar. This proves the statement. b) Fix p ∈ (0, 1) and a rational t ∈ ( 1 2 , 1). Choose a pair of positive integers m and l such that mt = l(1 −t) and set n = m + l. Let x i = t, i = 1, 2, . . . , m; x i = t − 1, i = m + 1, m + 2, . . . , n; y i = −1, i = 1, 2, . . . , m; y m+1 = m; y i = 0, i = m + 2, . . . , n. Then x ∈ R n 0 , max i x i − min i x i = 1, y ∈ Z n 0 and x p p − x + y p p = m(t p −(1 −t) p ) + (1 −t) p − (m −1 + t) p , which is possitive for m big enough. Problem 6. Suppose that F is a family of finite subsets of N and for any two sets A, B ∈ F we have A ∩ B = Ø. a) Is it true that there is a finite subset Y of N such that for any A, B ∈ F we have A ∩ B ∩ Y = Ø? b) Is the statement a) true if we suppose in addition that all of the members of F have the same size? Justify your answers. Solution. a) No. Consider F = {A 1 , B 1 , . . . , A n , B n , . . .}, where A n = {1, 3, 5, . . . , 2n− 1, 2n}, B n = {2, 4, 6, . . . , 2n, 2n + 1}. b) Yes. We will prove inductively a stronger statement: Suppose F , G are two families of finite subsets of N such that: 1) For every A ∈ F and B ∈ G we have A ∩B = Ø; 2) All the elements of F have the same size r, and elements of G – size s. (we shall write #(F ) = r, #(G) = s). 7 Then there is a finite set Y such that A ∪ B ∪ Y = Ø for every A ∈ F and B ∈ G. The problem b) follows if we take F = G. Proof of the statement: The statement is obvious for r = s = 1. Fix the numbers r, s and suppose the statement is proved for all pairs F  , G  with #(F  ) < r, #(G  ) < s. Fix A 0 ∈ F , B 0 ∈ G. For any subset C ⊂ A 0 ∪B 0 , denote F (C) = {A ∈ F : A ∩ (A 0 ∪B 0 ) = C}. Then F = ∪ Ø=C⊂A 0 ∪B 0 F (C). It is enough to prove that for any pair of non- empty sets C, D ⊂ A 0 ∪B 0 the families F (C) and G(D) satisfy the statement. Indeed, if we denote by Y C,D the corresponding finite set, then the finite set ∪ C,D⊂A 0 ∪B 0 Y C,D will satisfy the statement for F and G. The proof for F (C) and G(D). If C ∩D = Ø, it is trivial. If C ∩ D = Ø, then any two sets A ∈ F (C), B ∈ G(D) must meet outside A 0 ∪ B 0 . Then if we denote ˜ F (C) = {A \ C : A ∈ F (C)}, ˜ G(D) = {B \ D : B ∈ G(D)}, then ˜ F (C) and ˜ G(D) satisfy the conditions 1) and 2) above, with #( ˜ F (C)) = #(F ) − #C < r, #( ˜ G(D)) = #(G) − #D < s, and the inductive assumption works. 8 . which 1 + 1 n < θ k 1 (θ 0 = α) so that for each k, (1) 1 + 1 n k < θ k 1 ≤ 1 + 1 n k − 1 . Since θ k 1 ≤ 1 + 1 n k − 1 we have 1 + 1 n k +1 <. + 1 M 4  ··· = 1+ 1 M + 1 M 2 + 1 M 3 +··· = 1+ 1 M − 1 . Assume that for some k we have 1 + 1 n k − 1 < θ k 1 . Then we get α (1 + 1 n 1 ) (1 + 1 n 2 )